This is the first problem of three problems about a linear recurrence relation and linear algebra.

Proof.

(a) Find the eigenvalues and eigenvectors of $T$.

By problem 2, the matrix representation of the linear transformation $T: U\to U$ with respect to the basis $B=\{\mathbf{u}_1, \mathbf{u}_2\}$ of $U$ is given by
\[A=\begin{bmatrix}
0 & 1\\
-3& 5
\end{bmatrix}.\]
The eigenvalues of $T$ is the same as the eigenvalues of the matrix $A$.
The characteristic polynomial for $A$ is
\begin{align*}
p(t)=\det(A-tI)=\begin{vmatrix}
-t & 1\\
-3& 5-t
\end{vmatrix}=t^2-5t+3.
\end{align*}
Solving $p(t)=0$, the eigenvalues of $T$ are
\[t=\frac{5\pm \sqrt{13}}{2}.\]

Let us find an eigenvector corresponding to $t=\frac{5+ \sqrt{13}}{2}$.
If $(a_i)_{i=1}^{\infty}$ is an eigenvalue, then we have
\[T\big( (a_i)_{i=1}^{\infty}\big)=\frac{5+ \sqrt{13}}{2} (a_i)_{i=1}^{\infty}.\]
That is,
\begin{align*}
(a_2, a_3, a_4, \dots)=\frac{5+ \sqrt{13}}{2}(a_1, a_2, a_3, \dots).
\end{align*}
Thus, the eigenvector satisfies
\[a_{i+1}=\frac{5+ \sqrt{13}}{2}a_i,\]
hence it is a geometric progression with initial value $a_1$ and common ratio $\frac{5+ \sqrt{13}}{2}$. Thus, we have
\[(a_i)_{i=1}^{\infty}=\left( \left(\frac{5+ \sqrt{13}}{2} \right )^{i-1} a_1 \right)_{i=1}^{\infty}\]
is an eigenvector for any $a_1\in \R$.

Similarly, we see that the eigenvectors for $t=\frac{5- \sqrt{13}}{2}$ is
\[(a_i)_{i=1}^{\infty}=\left( \left(\frac{5- \sqrt{13}}{2} \right )^{i-1} a_1 \right)_{i=1}^{\infty}\]
for any $a_1\in \R$.
Since $U$ is a $2$-dimensional vector space, these are all the eigenvectors.

By the result of (a), we know that
\[\mathbf{v}_1=\left( \left(\frac{5+ \sqrt{13}}{2} \right )^{i-1} \right)_{i=1}^{\infty}\]
and
\[\mathbf{v}_2=\left( \left(\frac{5- \sqrt{13}}{2} \right )^{i-1} \right)_{i=1}^{\infty}\]
are eigenvectors of $t=\frac{5\pm \sqrt{13}}{2}$, respectively. (We took the initial value $a_1=1$.)
Thus, the set $B’=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis of $U$.

(c) Find the formula for the sequences $(a_i)_{i=1}^{\infty} \in U$.

Let $(a_i)_{i=1}^{\infty}\in U$. Using the basis vectors in $B=\{\mathbf{u}_1, \mathbf{u}_2\}$, we have
\[(a_i)_{i=1}^{\infty} = a_1 \mathbf{u}_1+ a_2\mathbf{u}_2.\]
Since $B’=\{\mathbf{v}_1, \mathbf{v}_2\}$ is also a basis for $U$, we can also write
\[(a_i)_{i=1}^{\infty} = b_1 \mathbf{v}_1+ b_2\mathbf{v}_2\]
for some scalars $b_1, b_2$.

We want to express $b_1, b_2$ in terms of $a_1, a_2$.
To do this, we first express $\mathbf{u}_1$ and $\mathbf{u}_2$ as linear combinations of $\mathbf{v}_1, \mathbf{v}_2$.

Let $\mathbf{u}_1=c_1\mathbf{v}_1+c_2\mathbf{v}_2$ for some $c_1, c_2$. We determine $c_1, c_2$.
Comparing the first two terms, we have
\begin{align*}
1&=c_1+c_2\\
0&=c_1 \left(\frac{5+ \sqrt{13}}{2} \right )+c_2 \left(\frac{5- \sqrt{13}}{2} \right ).
\end{align*}
Solving these equations, we have
\[c_1=\frac{13-5\sqrt{13}}{26}, c_2=\frac{13+5\sqrt{13}}{26}.\]
Thus we have obtained the linear combination
\[\mathbf{u}_1=\frac{13-5\sqrt{13}}{26} \mathbf{v}_1+\frac{13+5\sqrt{13}}{26}\mathbf{v}_2. \tag{*}\]

Similarly, if $\mathbf{u}_2=c_1\mathbf{v}_1+c_2\mathbf{v}_2$ for some $c_1, c_2$, then comparing the first two terms we have
\begin{align*}
0&=c_1+c_2\\
1&=c_1 \left(\frac{5+ \sqrt{13}}{2} \right )+c_2 \left(\frac{5- \sqrt{13}}{2} \right ).
\end{align*}
and this gives
\[c_1=\frac{\sqrt{13}}{13}, c_2=-\frac{\sqrt{13}}{13}.\]
Thus we have
\[\mathbf{u}_2=\frac{\sqrt{13}}{13} \mathbf{v}_1-\frac{\sqrt{13}}{13}\mathbf{v}_2. \tag{**}\]

Using (*), (**), we have
\begin{align*}
&(a_i)_{i=1}^{\infty} = a_1 \mathbf{u}_1+ a_2\mathbf{u}_2\\
&=a_1\left(\frac{13-5\sqrt{13}}{26} \mathbf{v}_1+\frac{13+5\sqrt{13}}{26}\mathbf{v}_2 \right)+a_2\left(\frac{\sqrt{13}}{13} \mathbf{v}_1-\frac{\sqrt{13}}{13}\mathbf{v}_2 \right)\\
&=\left(\frac{13-5\sqrt{13}}{26}a_1+\frac{\sqrt{13}}{13}a_2\right) \mathbf{v}_1
+\left( \frac{13+5\sqrt{13}}{26}a_1-\frac{\sqrt{13}}{13}a_2 \right) \mathbf{v}_2.
\end{align*}
The coefficients in the last linear combination are the $b_1, b_2$ that we wanted to find.

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